Current flow direction in BJT

Thread Starter

mau80

Joined Oct 24, 2010
21
Hi, I've found some problem finding out the voltage of o node in a very simple circuit. u node
I cannot find out how the current in emitter in a NPN BJT could have negative sign.

I'd like to ask your help in this analysis.

I've attached the circuit simulated by Pspice, and my computation.

Could you give to me some input?

Thank you very much for your help!

Maurizio
 

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Jony130

Joined Feb 17, 2009
5,487
Well the current is "positive", only simulation assume that current is "negative" when he comes out form the element terminal.

Vu = I2*R2 = ( Ie + (Vcc - Vu)/R3) * R2

Ie = (β+1) *Ib

Ib = (Vcc - Vu+Vbe)/R1

Or

I2 = I3 + Ie

I2 = Vu/R2

Ie = [(Vcc - (Vu+Vbe))/R1 ] * (β+1)

I3 = (Vcc - Vu)/R3

And if we solve this

Vu = (R2* (R1*Vcc - R3*(Vbe - Vcc) (1 + β)) / (R1*(R2 + R3) + R2*R3 (1 + β)) =

=( R1*Vcc - R3*(1 + β)*(Vbe - Vcc)) / ( R1 + R3 + (R1 R3)/R2 + R3*β ) = 2.5551V

Or we could use superposition for this circuit:




And then by inspection we can write

Vu = Vcc * ( R1/(β+1) || R2) / ( R3 + ( R1/(β+1) || R2) + (Vcc - Vbe) * R2||R3 / ( R1/(β+1) + R2||R3 )

And since R3>> R2 and R1 then

Vu ≈ (Vcc - Vbe) * R2||R3 / ( R1/(β+1) + R2||R3 ) ≈ (Vcc - Vbe) * R2 / ( R1/(β+1) + R2 ) = 2.54816514V
 

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Thread Starter

mau80

Joined Oct 24, 2010
21
Dear Jony,

thank you very much for your reply.
I'd like to ask if you can tell me how you've write following formula:

Vu = (R2* (R1*Vcc - R3*(Vbe - Vcc) (1 + β)) / (R1*(R2 + R3) + R2*R3 (1 + β)) =

=( R1*Vcc - R3*(1 + β)*(Vbe - Vcc)) / ( R1 + R3 + (R1 R3)/R2 + R3*β ) = 2.5551V

I can follow you solution until:

Vu = I2*R2 = ( Ie + (Vcc - Vu)/R3) * R2

Ie = (β+1) *Ib

Ib = (Vcc - Vu+Vbe)/R1

Or

I2 = I3 + Ie

I2 = Vu/R2

Ie = [(Vcc - (Vu+Vbe))/R1 ] * (β+1)

I3 = (Vcc - Vu)/R3


Thank you very much for your great help!

Bye
Maurizio
 

Jony130

Joined Feb 17, 2009
5,487
Have you try solve it by yourself ?

I2 = I3 + Ie

Vu/R2 = [(Vcc - (Vu+Vbe))/R1 ] * (β+1) + (Vcc - Vu)/R3

In this step it is end of the electronic, but the beginning of mathematics.
 

Thread Starter

mau80

Joined Oct 24, 2010
21
I've reached the correct result following your suggestion.

I'd like to ask last thing. Where is the error in the attached computation.
Why I don't find correct result?
I think taht Vu can be computed a Vcc - I3R3, right?
I've spent a lot of time trying to find out this but I've not found.

Thank you very very much!

Maurizio
 

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Thread Starter

mau80

Joined Oct 24, 2010
21
Thank yuo for your help. Finnally I was able to find out the error.
This error is in considering:

Vcc - Vu + G_gamma

The correct form is:

Vcc - ( Vu + V_gamma )

Vcc - Vu - V_gamma

This give correct result.

Thank you very much for your suggestion and help!

Bye
Maurizio
 
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